I want to prove that if $p(x)\in F[x]$, where $F$ is a field, is irreducible and $F$ has characteristic zero, then $p(x)$ has no repeated roots.
I found this argument in a book, but I don't understand something:
If $p(x)$ is irreducible and has a repeated root, then $p'(x)=0$. Now if $F$ has characteristic zero, this implies $p(x)$ is constant and the claim follows.
It is easy to see that $p'(x)=0$ implies $p(x)$ is constant, since $F$ has characteristic zero.
What I don't understand is why if $p(x)$ is irreducible and has a repeated root, then $p'(x)=0$.
If $p(x)$ has a repeated root, then $p(x)=(x-a)^2p_1(x)$ where $a$ is the repeated root in its splitting field. Also, it is $p'(a)=0$. But why it implies $p'(x)=0$?
Thank you.
Best Answer
Suppose that $p'(x)$ is not identically $0$. If it has a root, then it has degree $\ge 1$.
The $\gcd$ of $p(x)$ and $p'(x)$ over $F$ cannot be $1$. For if it is $1$, there are polynomials $s(x)$ and $t(x)$ in $F[x]$ such that $s(x)p(x)+t(x)p'(x)=1$. But then $p(x)$ and $p'(x)$ cannot have a common root in any field.
So $\gcd(p(x),p'(x))\ne 1$, which contradicts the irreducibility of $p(x)$, since $p'(x)$ has degree less than the degree of $p(x)$.